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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th

If for some q, q, r $\in$ R, not at all have same sign, one of the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$ is also a root of the equation x² + 2x – 8 = 0, then $\frac{q^{2} + r^{2}}{p^{2}}$ is equal to………….

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Raj Pandey | Contributor-Level 9

2 weeks ago

Let a and b are the roots of $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$

$∴ α + β > 0 a n d α β > 0$ Also, it has a common root with x² + 2x – 8 = 0

$∴$ The common root between above two equations is 4.

$\Rightarrow 16 (p^{2} + q^{2}) - 8 q (p + r) + q^{2} + r^{2} = 0 \Rightarrow (16 p^{2} - 8 p q + q^{2}) + (16 q^{2} - 8 q r + r^{2}) = 0$

$\Rightarrow {(4 p - q)}^{2} + {(4 q - r)}^{2} = 0 \Rightarrow q = 4 p a n d r = 16 p ∴ \frac{q^{2} + r^{2}}{p^{2}} = \frac{16 p^{2} + 256 p^{2}}{p^{2}} = 272$

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If for some q, q, r $\in$ R, not at all have same sign, one of the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$ is also a root of the equation x² + 2x – 8 = 0, then $\frac{q^{2} + r^{2}}{p^{2}}$ is equal to………….

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If for some q, q, r ∈ R, not at all have same sign, one of the roots of the equation ( p 2 + q 2 ) x 2 − 2 q ( p + r ) x + q 2 + r 2 = 0 is also a root of the equation x2 + 2x – 8 = 0, then q 2 + r 2 p 2 is equal to………….

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If for some q, q, r $\in$ R, not at all have same sign, one of the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$ is also a root of the equation x² + 2x – 8 = 0, then $\frac{q^{2} + r^{2}}{p^{2}}$ is equal to………….